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Q.132 Chemistry ⭐ 360° Option Elimination Active
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NEET 2020 Chemistry Question 132 Solution

Topic: Chemical Kinetics · Official NTA Booklet Code E1
NEET 2020 Chemistry Question 132: First Order Kinetics: Time Required for 90% Reactant Reduction

Question

The rate constant for a first order reaction is 4.606×10⁻³ s⁻¹. The time required to reduce 2.0 g of the reactant to 0.2 g is:

✓
Verified Against NTA Final Answer Key
Option (3) : 500 s
📚 Chemical Kinetics Easy
📋 Given Parameters & Problem Setup
Reaction Order First order
Rate Constant (k) 4.606 × 10^-3 s^-1
Initial Amount [R]0 2.0 g
Remaining Amount [R] 0.2 g
Integrated Rate Formula t = (2.303 / k) × log([R]0 / [R])
⚡

The 30-Second Exam Hall Shortcut: ⚡ "Ratio [R]0/[R] = 2.0 / 0.2 = 10 -> log(10) = 1; t = 2.303 / (4.606 × 10^-3) = 500 s!"

⏱️ 25 seconds

The integrated rate equation for a first order reaction is t=2.303klog⁡([R]0[R])t = \frac{2.303}{k} \log\left(\frac{[R]_0}{[R]}\right). Here, the concentration ratio is 2.00.2=10\frac{2.0}{0.2} = 10, and log⁡(10)=1\log(10) = 1. Therefore, t=2.3034.606×10−3×1=12×10−3=10002=500 st = \frac{2.303}{4.606 \times 10^{-3}} \times 1 = \frac{1}{2 \times 10^{-3}} = \frac{1000}{2} = 500\text{ s}! Mark Option (3) in 10 seconds.

1 Formula: t = (2.303 / k) × log([R]0 / [R]).
2 Substitute concentrations: [R]0 = 2.0 g, [R] = 0.2 g.
3 Ratio: [R]0 / [R] = 2.0 / 0.2 = 10.
4 log(10) = 1.
5 Calculate: t = 2.303 / (4.606 × 10^-3) = 1 / (2 × 10^-3) = 500 s.
6 Select Option (3).
🛡️
Why You Can 100% Trust This Shortcut: NCERT Chemistry · Chemical Kinetics (Integrated Rate Equations: First Order Reactions)
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

State Integrated Rate Law for First Order Reaction

For a first order reaction R→ProductsR \to \text{Products}, the integrated rate equation in common logarithmic form is: k=2.303tlog⁡([R]0[R])  ⟹  t=2.303klog⁡([R]0[R])k = \frac{2.303}{t} \log\left(\frac{[R]_0}{[R]}\right) \implies t = \frac{2.303}{k} \log\left(\frac{[R]_0}{[R]}\right)

Step 2

Insert Given Quantities

Given: - Rate constant k=4.606×10−3 s−1k = 4.606 \times 10^{-3}\text{ s}^{-1} - Initial mass of reactant [R]0=2.0 g[R]_0 = 2.0\text{ g} - Final mass of reactant remaining [R]=0.2 g[R] = 0.2\text{ g} (Notice that for a given volume, the ratio of masses is identical to the ratio of molar concentrations).

Step 3

Evaluate the Logarithmic Term

[R]0[R]=2.00.2=10\frac{[R]_0}{[R]} = \frac{2.0}{0.2} = 10 log⁡(10)=1\log(10) = 1

Step 4

Calculate Time (t)

t=2.3034.606×10−3×1t = \frac{2.303}{4.606 \times 10^{-3}} \times 1 Notice that 4.606=2×2.3034.606 = 2 \times 2.303: t=12×10−3=10002=500 secondst = \frac{1}{2 \times 10^{-3}} = \frac{1000}{2} = 500\text{ seconds} This confirms Option (3).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan
1 A
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; arithmetic error in dividing 1000 by 2.

⚠️
Student Fallacy (Why candidates pick this trap): Dividing by 10 instead of 2.
2 B
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; inverts the factor of 2 or forgets to account for log⁡(10)\log(10).

⚠️
Student Fallacy (Why candidates pick this trap): Multiplying by 2 instead of dividing.
3 C
✓ 100% Correct
Verdict & Scientific Reason:

Factually and mathematically exact; t=2.3034.606×10−3=500 st = \frac{2.303}{4.606 \times 10^{-3}} = 500\text{ s}.

💡
Why this is airtight: None; clean cancellation of 2.303 into 4.606.
4 D
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; neglects the factor of 2 (4.606/2.303=24.606 / 2.303 = 2) in the denominator.

⚠️
Student Fallacy (Why candidates pick this trap): Forgetting that 4.606 is double of 2.303.
⚠️

NTA Trap Alert: Numerical Cancellation Landmark

NTA intentionally uses k=4.606×10−3k = 4.606 \times 10^{-3} because 4.606/2.303=24.606 / 2.303 = 2 exactly! Do not waste time doing long division by hand.

🛡️ How to avoid the -1 mark: Spot the multiple: 2.303/4.606=1/22.303 / 4.606 = 1/2!
📖

Core NEET Syllabus Standard

Chemistry · Chemical Kinetics Standard Benchmark
Core Scientific Principle Tested:

""

Syllabus Focus: First Order Kinetics: Time Required for 90% Reactant Reduction
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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